Mechanics question (beginner level)

[trew]

Homework Statement

Find acceleration of the box

Relevant Equations

coefficient of friction and

I'm okay with the concept of resolving into two components.

BUT, with re: to resolving perpendicular to the acceleration, I don't understand how 25cos80 comes into?

Surely it should be included when I'm resolving in the direction of the acceleration?

 

[Doc Al]

Don't forget that cos(80) = sin(10). Sine and cosine are co-functions.

 

[haruspex]

Surely it should be included when I'm resolving in the direction of the acceleration?

The component of the 25N in the direction of acceleration would be 25 cos(10). This is the other component.

 

[trew]

Don't forget that cos(80) = sin(10). Sine and cosine are co-functions.

The component of the 25N in the direction of acceleration would be 25 cos(10). This is the other component.

Thanks for your help so far but tbh I'm still lost.

Here's what I've done so far:

So I've resolved the first part into 2gsin10 and 2gcos10 but after looking at the problem for the past hour I still can't see how I can break down the 25N into two components.

 

[Doc Al]

I still can't see how I can break down the 25N into two components.

Draw a similar triangle for that 25N force. The force acts horizontally: you want its components perpendicular and parallel to the incline. What angle does the force make with the incline?

 

[trew]

Draw a similar triangle for that 25N force. The force acts horizontally: you want its components perpendicular and parallel to the incline. What angle does the force make with the incline?

Ok so I did that and this is what I got:

so I can't see how I would still get 25cos10 for the hypotenuse side? Or have I done this triangle wrong?

 

[Doc Al]

Or have I done this triangle wrong?

Yes, the triangle is wrong. The 25N force must be the hypotenuse of the right triangle. (The components will be the shorter sides.)

 

[trew]

Yes, the triangle is wrong. The 25N force must be the hypotenuse of the right triangle. (The components will be the shorter sides.)

Ok I'm back a day later and I still don't get it

I looked at this video:
which has pretty much the same question and I still can't visualise the triangle with the 25N force being the hypotenuse.

I've tried many combinations but I can't form a triangle with the 25N and it's two components.

 

[trew]

Yes, the triangle is wrong. The 25N force must be the hypotenuse of the right triangle. (The components will be the shorter sides.)

Also, I like using sin0=opp/hyp and cos0=adj/hyp to work out these forces but since I can't see how that triangle is formed I can't use this method.

For me it's the best method since I can learn what's going intuitively as opposed to assuming that cos is associated with the x-direction and sin with the y-direction.

 

[Doc Al]

Also, I like using sin0=opp/hyp and cos0=adj/hyp to work out these forces but since I can't see how that triangle is formed I can't use this method.

Nothing wrong with using a right triangle to find the components. Here's a diagram for a similar problem (that I found on the web). Perhaps it might inspire you:

In that diagram, the applied force Fa is horizontal just like in your problem. (See if you can find the correct right triangle.)

For me it's the best method since I can learn what's going intuitively as opposed to assuming that cos is associated with the x-direction and sin with the y-direction.

Good! Never do things blindly as you can easily make a mistake.

Nonetheless, as long as the angle Θ is with respect to some axis (whatever that is in your problem), then FcosΘ will be the component parallel to that axis.

 

[Herman Trivilino]

Can you find the x and y components of the 25 N force?